Rational Root-Rational Zero Theorem worksheet - Free Printable

I will solve this problem by applying the Rational Zero Theorem to each polynomial. The theorem states that for a polynomial with integer coefficients, any possible rational zero must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.

Let me analyze each polynomial one by one.

First Polynomial: 2x³ - x² + 19x + 10

Step 1: Identify the constant term and leading coefficient
- Constant term (a₀) = 10
- Leading coefficient (aₙ) = 2

Step 2: Find all factors of the constant term (p)
Factors of 10: ±1, ±2, ±5, ±10

Step 3: Find all factors of the leading coefficient (q)
Factors of 2: ±1, ±2

Step 4: Generate all possible p/q values
Possible rational zeros: ±1/1, ±2/1, ±5/1, ±10/1, ±1/2, ±2/2, ±5/2, ±10/2
Simplifying: ±1, ±2, ±5, ±10, ±1/2, ±1, ±5/2, ±5
After removing duplicates: ±1, ±2, ±5, ±10, ±1/2, ±5/2

Looking at the green cards provided for this polynomial:
±1, ±2, ±3, ±4, ±5, ±10, ±1/2, ±2/5, ±5/2, ±5/3

Comparing with our calculated possible rational zeros:
- Possible: ±1, ±2, ±5, ±10, ±1/2, ±5/2
- Not possible: ±3, ±4, ±2/5, ±5/3

Second Polynomial: x⁵ + 3x⁴ - 18x² + 27

Step 1: Identify the constant term and leading coefficient
- Constant term (a₀) = 27
- Leading coefficient (aₙ) = 1

Step 2: Find all factors of the constant term (p)
Factors of 27: ±1, ±3, ±9, ±27

Step 3: Find all factors of the leading coefficient (q)
Factors of 1: ±1

Step 4: Generate all possible p/q values
Possible rational zeros: ±1/1, ±3/1, ±9/1, ±27/1
Simplifying: ±1, ±3, ±9, ±27

Looking at the green cards provided for this polynomial:
±1, ±2, ±3, ±4, ±5, ±6, ±9, ±27, ±1/3, ±1/9

Comparing with our calculated possible rational zeros:
- Possible: ±1, ±3, ±9, ±27
- Not possible: ±2, ±4, ±5, ±6, ±1/3, ±1/9

Third Polynomial: 5x³ - 23x² - 65x - 22

Step 1: Identify the constant term and leading coefficient
- Constant term (a₀) = -22
- Leading coefficient (aₙ) = 5

Step 2: Find all factors of the constant term (p)
Factors of 22: ±1, ±2, ±11, ±22

Step 3: Find all factors of the leading coefficient (q)
Factors of 5: ±1, ±5

Step 4: Generate all possible p/q values
Possible rational zeros: ±1/1, ±2/1, ±11/1, ±22/1, ±1/5, ±2/5, ±11/5, ±22/5
Simplifying: ±1, ±2, ±11, ±22, ±1/5, ±2/5, ±11/5, ±22/5

Looking at the green cards provided for this polynomial:
±1, ±2, ±3, ±4, ±5, ±10, ±11, ±12, ±22, ±1/5, ±2/5, ±11/5, ±22/5, ±5/11, ±5/22

Comparing with our calculated possible rational zeros:
- Possible: ±1, ±2, ±11, ±22, ±1/5, ±2/5, ±11/5, ±22/5
- Not possible: ±3, ±4, ±5, ±10, ±12, ±5/11, ±5/22

## Final Answer:

For Polynomial 1 (2x³ - x² + 19x + 10):
- POSSIBLE Rational Zeros: ±1, ±2, ±5, ±10, ±1/2, ±5/2
- NOT possible Rational Zeros: ±3, ±4, ±2/5, ±5/3

For Polynomial 2 (x⁵ + 3x⁴ - 18x² + 27):
- POSSIBLE Rational Zeros: ±1, ±3, ±9, ±27
- NOT possible Rational Zeros: ±2, ±4, ±5, ±6, ±1/3, ±1/9

For Polynomial 3 (5x³ - 23x² - 65x - 22):
- POSSIBLE Rational Zeros: ±1, ±2, ±11, ±22, ±1/5, ±2/5, ±11/5, ±22/5
- NOT possible Rational Zeros: ±3, ±4, ±5, ±10, ±12, ±5/11, ±5/22